Proving the Range of $2 - (x + 1)^2$ What I've been taught is that a good way to prove the range of a function is by using the two statements $\infty < x \leq 0$ and $0 \leq x < \infty$ then working from there to reach the function's form. Doing this here leads to the range being ]-infinity,1] which obviously isn't true as 2 is clearly in the range.
My steps: (I'll use $\vee$ to separate the two statements until they're the same)
$(-\infty < x \leq 0) \vee (0 \leq x < \infty)$  
$(-\infty < x+1 \leq 1) \vee (1 \leq x+1 < \infty$  
$(1 \leq (x+1)^2 < \infty) \vee (1 \leq (x+1)^2 < \infty)$
$-\infty < -(x+1)^2 \leq -1$  
$-\infty < 2-(x+1)^2 \leq 1$
What am I doing wrong ? Is the method I've been taught wrong to begin with ?
 A: Hint: it is $$2-(x+1)^2=2-(x^2+2x+1)=-x^2-2x+1$$
then you can use that $$\sqrt{2}^2-(x+1)^2=(\sqrt{2}-x-1)(\sqrt{2}+x+1)$$
A: Better do this way
Hint: Use the minimum value of a perfect square, $(x+1)^2 $in this case.    
Answer:Minimum value of $2-(x+1)^2$ is given when $(x+1)^2$ is maximum.
Maximum value of $(x+1)^2\Rightarrow \infty$
So minimum value of $2-(x+1)^2\Rightarrow -\infty$
Maximum value of $2-(x+1)^2$ is given when $(x+1)^2$ is minimum.
Minimum value of $(x+1)^2=0$
So maximum value of $2-(x+1)^2=2$
Therefore range is $(-\infty,2]$
A: The mistake is $-\infty < x + 1 \leq 1 \implies 1\leq (x+1)^2 <\infty$. It's wrong because $x\mapsto x^2$ is neither strictly increasing nor strictly decreasing on $(-\infty, 1]$. The correct way to do it would be something like this:
\begin{align}-\infty < x + 1 \leq 1 &\implies -\infty < x + 1 \leq 0\ \text{or}\ 0 < x + 1 \leq 1\\
&\implies 0 \leq (x+1)^2 < \infty\ \text{or}\ 0<(x+1)^2\leq 1\\
&\implies 0 \leq (x+1)^2 < \infty.\end{align}
However, the easiest way to do this is just to write $$(x+1)^2 \geq 0\implies (x+1)^2 - 2 \geq -2\implies 2-(x+1)^2 \leq 2.$$
A: $$2-(x+1)^2=2-[x^2+2x+1]=-[x^2+2x-1]$$
From there, complete the square:
$$\to -[(x+1)^2-2]$$
For $(x+1)^2$, $\min=0$, $\max\to \infty$
Hence for $-[(x+1)^2-2], \min\to-\infty , \max=2$
Thus the range you seek is:$$2-(x+1)^2\le 2$$
A: $y=2-(x+1)^2$
$\implies x=\pm (2-y)^{1/2}-1$
$$-\infty \leq x \leq \infty$$
$$\implies -\infty \leq \pm (2-y)^{1/2}-1\leq \infty$$
$$\implies -\infty \leq |2-y|\leq \infty$$
$$\implies 0 \leq |2-y|\leq \infty$$
Then $-\infty\leq y \leq 2$, just take when $|2-y|=2-y$
A: Let $u=x+1$. Since $x\in(-\infty,\infty)$, we have $$\begin{align}u^2\in[0,\infty)&\implies -u^2\in(-\infty,0]\\&\implies 2-u^2\in(-\infty,2]\\&\implies\color{red}{2-(x+1)^2\in(-\infty,2]}\end{align}$$ as required.
